Will Cadmium hydroxide precipitate from 0.01 M solution of #"CdCl"_2# at pH 9. #K_(sp)# of #"Cd"("OH")_2 = 2.5 xx 10^(-14)"M"^3#? Please solve,thanks!

1 Answer
Mar 23, 2016

Answer:

Yes, a precipitate will form.

Explanation:

Your strategy here will be to calculate the concentrations of the cadmium(II) cations, #"Cd"^(2+)#, and of the hydroxide anions, #"OH"^(-)#, and figure out if these concentrations are high enough to ensure that a precipitate is formed.

So, you know that the pH of the solution is equal to #9#. As you know, an aqueous solution at room temperature has

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " POH" = 14)color(white)(a/a)|)))#

This means that the pOH of the solution, which can be used to find the concentration of hydroxide anions, will be equal to

#"pOH" = 14 - 9 = 5#

This means that you have

#["OH"^(-)] = 10^(-"pOH") = 10^(-5)"M"#

Cadmium chloride, #"CdCl"_2#, is a soluble ionic compound that dissociates completely in aqueous solution to form cadium(II) cations and chloride anions

#"CdCl"_text(2(aq]) -> "Cd"_text((aq])^(2+) |+ 2"Cl"_text((aq])^(-)#

Notice that one mole of cadmium chloride produces one mole of cadmium cations in solution. This means that you have

#["Cd"^(2+)] = ["CdCl"_2] = 10^(-2)"M"#

The partial dissociation of cadmium hydroxide in aqueous solution looks like this

#"Cd"("OH")_text(2(s]) rightleftharpoons "Cd"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#

By definition, the solubility product constant, #K_(sp)#, for cadmium hydroxide will be equal to

#K_(sp) = ["Cd"^(2+)] * ["OH"^(-)]^color(red)(2)#

Now, in order to determine if a precipitate if formed or not, you need to calculate the reaction quotient, #Q_(sp)#.

The difference between #K_(sp)# and #Q_(sp)# lies in the fact that the former is calculated using equilibrium concentrations, while the latter is calculated using the concentrations of the ions at a given time.

In other words, #Q_(sp)# can be used to determine if a precipitate will form. In order for a precipitate to form, you need to have

#Q_(sp) > K_(sp)#

At this point, a precipitate will form until you reach saturation, i.e until #Q_(sp) = K_(sp)#.

If #Q_(sp) < K_(sp)#, the solution is unsaturated and a precipitate will not form.

Plug in the concentrations of the two ions to get

#Q_(sp) = 10^(-2)"M" * (10^(-5))^color(red)(2)"M"^color(red)(2) = 10^(-12)"M"^3#

Since you have

#10^(-12)"M"^3 > 2.5 * 10^(-14)"M"^3#

cadmium hydroxide will indeed precipitate out of solution.